motor fundamentals

7/24/2019 Motor Fundamentals

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ComprehensiveGuide to

UnderstandingMotorFundamentalsInstruction ManualII. INTRODUCTION. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 1II. MOTOR BASICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 1Stator Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1Rotor Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-2Simplified Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-3Rotor Current and Slip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-4Te !or"in# Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-$Tor%ue &s. Stator 'oles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-$Rotatin# Ma#netic (ields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-)Stator 'oles &s. Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-*Speed of Rotatin# (ield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1+Motor Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1+Te Rotor ,nder oad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-12i# (re%uenc/ Rotor Currents . . . . . . . . . . . . . . . . . . . . . . . . . 2-13Tor%ue Caracteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-140MA esi#n Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1

Te Motor as a Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-2+Motor osses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-21ffects of &olta#e &ariations . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-22Reduced &olta#e Startin# . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-2Motor Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-2$&olta#e ,n5alance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-2)ffects of (re%uenc/ &ariations . . . . . . . . . . . . . . . . . . . . . . . . . 2-2*

Ill. MOTORS AND ADJUSTABLE FREQUENCY DRIVES . 3 - 1'erformance Ad6anta#es . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-Operation A5o6e Base Speed . . . . . . . . . . . . . . . . . . . . . . . . . . 3-$Constant &olta#e Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-$Constant Tor%ue Ran#e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-1+7enerator Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-13

IV. ENERGY EFFICIENT MOTORS . . . . . . . . . . . . . . . . . 4-1

V. SPECIAL MOTORS . . . . . . . . . . . . . . . . . . . . . . . . . 5. - 1

II. MOTOR BASICSThe stator electromagnets are comprised of enameled wire


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wound over a permeable iron core. A typical motor may have 6,128

or 18 individual windings placed on a common core. The core is a

collection of stamped disks with slots for mounting the wire

windings (seeFigure 2-l).Stamped Steel is"s

STATORCOR!indin#sCannels (illedROTOR CORnum

Figure 2 - 1

Figure 2-2shows a simplified stator with only one set of

windings, labeled Al and A2. They are wound identically. If direct

current flows through Al and A2, they will be polarized north and

south as shown inFigure2-3(see page2-2). Notice that flux

crosses the gap between Al and A2, where the rotor is normally

mounted.Figure 2-2

2-1

AlFigure 2-3

The rotor itself has no wire windings, but does consist of

steel disks (or laminations), similar to the stator. The laminations

are perforated near the circumference (refer toFigure2-1, page

2-l).These disks are stacked together so that the perforations

align to form channels near the surface of the core assembly.

Through a careful die-casting process, the channels are then filledwith aluminum and a ring is formed on each end of the rotor. The

result is a row of conductive bars embedded into the rotor surface.

All of the bars are connected at each end of the rotor to form a

closed circuit for current flow (seeFigure 2-4).Rotor Barsnd Rin#s

Figure 2-4

Although the rotor bars are in direct contact with the steel

laminations, their resistance is much lower. Practically all rotor

current therefore flows in the bars, not in the laminations. Some

rotor designs employ a copper-brass alloy instead of aluminum.

This creates a lower resistance rotor circuit and thereby changes

motor performance. Sin-ce there are no wire windings, the squirrel

cage rotor is very rugged.2-2

To establish current flow in the rotor, there must first be a

voltage present on the rotor bars. This voltage is supplied by the


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magnetic field created by stator poles Al and A2. Simply

mounting the rotor in the gap between Al and A2 will not provide

voltage for the rotor circuit. However, if there was relative motion

between the rotor conductors and the stator field, a voltage would

be induced in the rotor bars (hence the name induction motor).

ie rFigure2-5 is a simplified view of an induction motor. It

details only two pairs of rotor bars (with end rings) and one set of

stator poles, labeled Al and A2. For clarity, Al and A2 are

permanent magnets instead of electromagnets. The motors shaft

is represented by the rotors centerline. In a squirrel cage motor,

the rotor voltage is induced by moving the stator's magnetic field

past the rotor bars. The rollers and hand crank inFigure 2-5

merely illustrate a means of rotating the magnetic field.

Figure 2-5

If the crank is turned while the rotor is at rest, the relative

motion between field and conductor will induce a voltage in the

rotor bars. Current will flow, producing a magnetic field around

each rotor bar as shown inFigure2-6a(see page2-4). The rotor

is attracted to the stator and begins to follow along in the same

direction. As the stator turns farther; the rotor bars pass out of

the magnetic field and the induced rotor voltage drops to zero (see

Figure2-6b, page2-4).Rotor current decays, as does the

resultant rotor field.

Continued rotation of the stator field produces the

conditions shown inFigure2-6c(see page2-4). Notice that the

stator has turned about 180 and now its magnetic field is cutting

the rotor bars again.2-3

Te !or"in# MotorThe model of the squirrel cage motor shown inFigure l-l

(see page l-l) is for illustration only. It is far from a practical

motor. An actual rotor has many bars, which are normally skewedat a slight anglelike Figure 2-8a.ComparingFigure 2-8aand

Figure 2-8bdepicts the difference between a skewed and

unskewed rotor.Figure 2-8a

Rotor 9it s"e9Figure 2-8b

Rotor 9itout s"e9Figure 2-8


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This prevents cogging (torque pulsations), which occurs if

the rotor bars are parallel to the stator poles. A normal squirrel

cage induction motor does not employ a crank to rotate the field,

either. The stator field rotates when 3~ alternating current is

applied to the stator.

If alternating current is applied to the single winding

shown inFigure2-2(see page2-1), the magnetic polarity will

reverse as the current reverses. This results in each pole swinging

north-south at the same frequency as the applied voltage. The

stator field can be considered to be rotating. A single phase

induction motor operates in this manner. However, this single

phase circuit has no starting torque. An extra starting circuit is

required to initially set the rotor in motion. If there is a 30

circuit, no extra starting circuit is required with the squirrel cage

motor.

Tor%ue &s. Stator 'olesFigure2-9ashows the stator poles of a 30 2 pole motor.

Two poles are energized by each of the three incoming phases. A 4

pole, 30 motor has twice as many stator poles per phase and is

illustrated inFigure2-9b.A motor may be constructed with any

even number of stator poles. The 4 pole motor is the most

commonly used. The 2, 6, and 8 pole motors are also popular.

Notice that the 4 pole stator has a flux pattern that contacts the

rotor in four places instead of only two(see Figure2-9c).Thisresults in twice the magnetic interaction between rotor and stator,

producing twice the amount of turning force (torque) at the motor

shaft.2-6

often operated from square waves generated by adjustable

frequency drives instead of sine waves of current. At low

frequencies (below 20 Hertz), a square waves magnetic field does

not shift smoothly from one position to the next, and cogging may

be noticeable.

Follow diagrams ofFigure2-10through times 3, 4, 5, 6

and 7 to verify that the magnetic field does, in fact, rotate. The

field at time 7 is identical to the field at time 1 in the cycle. The

magnetic field of a 2 pole motor makes one complete rotation

during one electrical cycle. If a rotor is placed inside the stator, it


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will make almost one rotation in one electrical cycle (depending on

the amount of slip between rotor and stator). The direction of

motor rotation may be reversed by switching any two of the stator

leads. This causes the stator field to reverse direction.

If the stator is wound with 4 poles per phase instead of two,

it might be compared to two separate 2 pole stators together.

Figure 2-11is a means to visualize this. Simply slide all six of the

stator poles onto one side of the core. When one cycle of 3 0 AC is

applied, the magnetic field rotates around only one-half of the

stator. The addition of six more poles on the right half of the stator

would allow a complete rotation to be made. However, it would

also require one more cycle of 3 0 power. A 4 pole stator field

rotates only once for each 2 cycle of applied current. Slide all 12 of

these stator poles onto one half of the core to create one half of an 8

pole motor. Logically, this motor demands 4 cycles of alternating

current to complete one rotation of its magnetic field. Also, an 8

pole stator is physically larger and draws more current than the 4

and 2 pole motors for the same HP or kW rating.

Concept of a 4 'ole StatorFigure 2-11

If the current that excites the 2, 4 and 8 pole motor is the

same frequency, then the 4 pole motor will run at one half the

speed of a 2 pole motor. The 8 pole motor will run at one-half the

speed of a 4 pole machine or one-fourth the speed of the 2 pole

motor.2-9

Speed of Rotatin# (ieldSince the stator field speed is exactly dependent on the

frequency of the applied current, it is said to be in synchronism

with the applied current.

The speed at which the stator field rotates is therefore

calledSynchronous Speed.The speed of the rotor is called

Running Speed.It differs only by the amount of slip as shown in

the equation below:

Where:

Ns=Synchronous speed of motor in RPMF = applied freuency in hert!P = num"er of poles per phase12# = con$ersion factor

(Equation 2-1)

Synchronous speed may be altered by changing either the

frequency applied or the number of poles. Multi-speed motors


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have external connections that allow an operator to switch the

stator from 2 poles to 4 poles or 4 poles to 6 poles, etc. This

provides only a limited number of definite speeds, such as those

shown inTable2-A(page2-7). This table lists the synchronous

speeds of various motors excited by the same frequency. If speeds

other than those listed are desired, it is necessary to change the

motors applied frequency.

This is commonly done by powering the motor from an

adjustable frequency drive. Using a drive to generate the applied

frequency produces an infinitely variable speed range from 0 RPM

to as high as 100,000 RPM (if required).

otor AccelerationLet us assume that the stator field is rotated at 1800 RPM

by 60 Hertz AC applied to the windings. Before the rotor begins to

turn, there is a slip of 1800 RPM. Relative motion between fieldand conductors is maximum, inducing a very high voltage in the

rotor. Rotor current is maximum and a strong magnetic field

results.Figure2-12plots the motor current in relation to rotor

speed.

The rotor accelerates from rest (point A), following the

stator. Slip decreases as the rotor accelerates. A rotor speed of

500 RPM (point B) means a slip of 1300 RPM. It also means less

relative motion between field and conductor and therefore a lower

induced voltage with less rotor current and a weaker rotor field.The rotor continues to accelerate, drawn by the stator field. When

the rotor reaches 1795 RPM (point E), its field is very weak. The

rotor bars are cutting very few lines of flux in a given time period.

If the rotor continued to accelerate, its current (and magnetic field)2-1#

otor ,nder oadIf no work is required at the motor shaft (no load applied),

the rotor will turn at approximately 1795 RPM (point E). Slip is

minimal and the resulting rotor field is very weak. Yet, rotorstator

attraction is just strong enough to produce enough torque to

keep the rotor turning at 1795 RPM. If a load is now applied to

the motor shaft, it immediately slows down, since the torque being

produced was just enough to keep the rotor at 1795 RPM. If the

stator speed remains constant (1800 RPM), then as the rotor slows

down, its bars automatically begin to cut more lines of stator flux


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per unit of time. The rotor field strengthens and more torque is

produced. The rotor will continue to slow down (or slip back) until

adequate torque is produced to power the load and maintain

rotation. This might correspond to point D inFigure2-12(see

page2-11). where slip has increased to 50 RPM, and current has

increased to 100%. If the load is removed, the torque being

produced will cause the rotor to accelerate back to its no load speed

of 1795 RPM (point E).

If the load is increased, again the motor will slow down to

produce greater torque by the rotor bars cutting more lines of flux.

Speed stabilizes when the motor torque matches the load torque,

this time perhaps at point C. The motor will maintain this speed

as long as the load is constant. Further load increases cause

greater current flow; beyond the safe limits of motor operation. An

overpowering load may cause the motor to stall completely, causing

extremely high currents to flow (point A). Overcurrent protection

must be installed to protect the motor in such an event. Under

normal circumstances this magnitude of current occurs only

during starting, (which lasts less than one second) while the rotor

attains its running speed.

When the rotor is stationary, either when stalled by a load

or during starting, rotor frequency is equal to stator frequency. If

the stator is excited by 60 Hz, then rotor frequency is also 60 Hz or

if the stator is excited by 50 Hz the rotor frequency is also 50 Hz.This is true regardless of the number of stator poles a motor has.

A 2 pole stator turns at 3600 RPM (@ 60 Hz) or 3000 RPM (@ 50

Hz) and induces 1 cycle of rotor voltage as each pole pair passes a

rotor bar. A 4 pole sator turns at half the speed but has twice the

pole pairs cutting each rotor bar. It induces 2 cycles of rotor

voltage per revolution, matching the rotor frequency of a 2 pole

machine.2-12

When the rotor field shifts (during high slip) it not only

affects motor current, but torque as well. Recall that torque

results from the magnetic attraction between rotor and stator.

Increased inductive reactance causes the rotor current (and

resultant field) to lag rotor voltage. Since the rotor voltage is

in-phase with the stator field, the rotor field must be out of phase

with the stator field.Figure2-13ashows the rotor field lagging


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the stator field and rotor voltage by the angle m (the cosine of x is

a measure of power factor). Notice that positive motor torque is

produced only during the periods when stator and rotor fields are

in-phase. Compare this withFigure2-13b,where voltage and

current are both in-phase. The out-of-phase rotor field(in

Figure2-13a)actually produces negative torque, or a retarding

force. As power factor decreases further, torque suffers more.

Figure2-13cshows a condition of maximum current lag ( m - 90>

resulting in negative torque completely cancelling positive torque.

Although maximum voltage and current flow in such a circuit, no

power is produced.

The torque equation of a squirrel cage induction motor is

similar to that for a DC shunt motor as shown in the following

equations. The major difference is the term cos m ". This

stipulates that only the in-phase rotor current produces positive

torque. Inductive reactance causes this phase shift, so it would be

expected that motor torque would be worst when inductive

reactance is greatest.%& Shunt Motor 'orue = ()# 1)

Where:(' = Motor 'orue &onstant# = *ield *lu+, = rmature &urrent

(Equation 2-2)

,nduction Motor 'orue = () ,)cos m

Where:KT = Motor 'orue &onstant/ = *ield *lu+,0 = Rotor &urrentcos c-c = Phase %isplacement of Rotor &urrent

(Equation2-3) 2-3)

2-1

As acceleration continues, rotor frequency and inductive

reactance decreases. The rotor flux moves more in-phase with

stator flux and torque increases. Maximum Torque (or

Break-down Torque) is developed at point C inFigure2-14,where

inductive reactance becomes equal to the rotor resistance. Beyond

point C, (points D, E and F) the inductive reactance continues to

drop off, but rotor current also decreases at the same rate,

reducing torque.

stator

Point G is synchromous speed and proves that if rotor and


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are at the same speed, rotor current and torque are zero.

At running speed, the motor will operate between points F

and D, depending on load. However, temporary load surges may

cause it to slip all the way back near point C on the knee of the

curve.

Beyond point C, the power factor decreases faster than

current increases, causing torque to drop-off. On the linear part of

the motor curve (points C to G), rotor frequency is only 1 to 3 Hertz

- almost DC. Inductive reactance is essentially zero and rotor

power factor approaches unity. Torque and current now become

directly proportional - 100% current produces 100% torque. If a

motor has a nameplate current of 3.6 amps, then when it draws 3.6

amps (at proper voltage and frequency) it must be producing 100%

of its nameplate torque. Torque and current remain directly

proportional up to approximately 10% slip. This relationship isvery useful when troubleshooting the motor and driven machine.

Notice that as motor load increases from zero (point F) to

100%, (point E) the speed drops only 45-55 RPM, about 3% of

synchronous speed. This makes the squirrel cage induction motor

very suitable for most constant speed applications (such as

conveyors) where, in some cases, 3% speed regulation might be

acceptable. This compares favorably with shunt wound DC

motors. If better speed regulation is required, the squirrel cage

motor may be operated from a closed loop regulator. As an

alternative, a reluctance synchronous induction motor (see pages

2-12 and 2-13) may be used instead of a squirrel cage induction

motor.

0MA esi#n ClassesIn the U.S., the National Electrical Manufacturers

Association (NEMA) has a standard on motors and generators;

NEMA Standards Publication No. MG 1. This standard has

classified squirrel cage motors according to their locked rotor

torque and current, breakdown torque, pull-up torque, and percent

slip. The four major classifications are A, B, C and D. They differ

primarily in the amount of rotor resistance and inductance each

possesses. Increasing a rotors resistance increases Starting

Torque but also decreases Breakdown Torque, overall efficiency,

and speed regulation (slip increases).

2-17


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The NEMA design D motor has the highest resistance rotor

for maximum starting torque(Figure2-18, seepage2-19). It has

no Pull-up or Breakdown Torque and is used where high starting

torque is critical. Acceleration of the flywheel on a punch press is

an excellent application for NEMA D motors.

These motors have the lowest efficiency of the 4 types

mentioned. They are available in two ranges of slip: 5% to 8% and

8% to 13%. The higher slip motors produce greater Starting

Torque, but lower slip motors possess a greater instantaneous

overload capability.

Figure 2-19shows the phase to phase wiring of a standard

squirrel cage motor. Note the resemblance to a 3 phase

transformer. The stator is the primary and the rotor acts as the

secondary. This comparison helps to emphasize the minimalmaintenance required on squirrel cage motors, Its just a

transformer that rotates. How much maintenance do you perform

on transformers?

1''ase A

$ STATORROTORTe Motor is 6er/ similar to a Transformer

Figure 2-19

2-2#

A transformers magnetic coupling is fixed, but the motors

coupling between stator and rotor changes. When slip is 1, the

coupling is optimum. This occurs at locked rotor, when the

induced rotor voltage is highest and rotor current is maximum,

Figure2-12, point A(page2-11). This may be compared to the

transformer with a shorted secondary circuit. As the rotor

accelerates, slip decreases from 1 to as little as 0.1 (no load speed).

The primary to secondary coupling deteriorates and induced rotor

voltage is very low. Rotor current is as low as the secondary

current in an unloaded transformer.

Although an unloaded transformer may draw only 2% of

rated primary current,Figure2-12, point E(page2-11) shows

that the unloaded squirrel cage stator draws about 40% current.

One reason for this is because the motors magnetic coupling is not

as efficient as a transformer coupling. This is due to the air gap


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between stator and rotorFigure2-9(page2-7). The stator must

draw extra current to increase the flux density of its magnetic field

and bridge the air gap. A narrow air gap requires less no-load

current than a wide air gap, so motor designs incorporate the

narrowest air gaps possible. This improves efficiency and power

factor.

However, if the air gap is too small there is a danger that

the rotor may actually contact the stator, short circuiting it.

Maintaining a precise air gap requires precision rotor bearings and

careful machining.

There are other no-load losses besides the air gap. The

rotor must produce enough torque to spin the shaft-mounted fan to

cool itself. This loss is known as Windage. The rotor must also

overcome friction from the bearings which support it. There are

also iron (Hysteresis and Eddy Current) and copper (12R> losses to

consider. The sum of these losses result in the unloaded motor

drawing as much as 40% of rated current. It is more important,

therefore, to carefully size induction motors. An oversized motor

might never operate at rated horsepower, resulting in poor

efficiency and power factor.

Figure2-20(see page2-22) shows the relationship

between efficiency, power factor and load for an average squirrel

cage motor. A fully loaded motor is more efficient than the same

motor running at quarter load. Also, a large HP motor is moreefficient than a small motor. High speed motors are more efficient

than low speed motors. Medium and low voltage motors (230 to

460V) are more efficient than high voltage (2300V) motors and

normal slip motors (NEMA A, B, C) are more efficient than high

slip (NEMA D) motors.

2-21

100

85

((ICI0C:0 25

'o9er (actorIllI IIll III1

75 100 125

OA ;Figure 2-20

ffects of &olta#e &ariationsRecall fromEquation2-3that torque is determined by

stator flux and rotor current, which result from rotor voltage. For

a given slip, rotor voltage is directly proportional to the density of


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the stator flux being cut. Stator flux is directly proportional to

stator current (until the saturation level is reached). Therefore, if

the motors applied frequency is held constant, stator current and

flux will rise and fall in direct relation to the applied voltage.

Increased voltage yields higher flux density, which means

that higher rotor voltages are induced at a given slip.

It follows then that less slip is required to produce rated

torque at the motor shaft. The slip from no-load to 150% full load

becomes more vertical as shown inFigure 2-21,where stator

applied voltage has been increased by 10%. Notice that the entire

torque curve is affected by this increase.;O(

9,?1221++)12S'

Figure 2-21

2-22

If applied voltage is decreased, flux is decreased, flux and

the rotor must slip back more in order to produce rated torque.

This is clear from the 90% and 50% curves inFigure2-21.Also

observe that 90% rated voltage, does not produce 90% of the rated

torque, but rather only 81%. 50% voltage only produces 25% rated

torque. This is because both stator flux and rotor current are

affected by voltage variations. As such, torque varies not directly,

but rather by the square of the applied voltage. With some

manipulation (not elaborated here),Equation2-3may be restated

from the voltage standpoint:

Where:T = torue at any slipKt = a ne torue constant = $olta4e applied to the stator indin4s

(Equation 2-4)

The relationship between voltage and torque is especially

important during troubleshooting. Low voltage can cause any of

the following problems:

Inadequate Starting Torque- If a motor and load are

closely matched, then a 10% drop in line voltage (even momentary)

during starting could result in load torque demand exceeding


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motor torque.

Speed Fluctuations- A momentary drop in line voltage

will cause a proportional dip in speed. Compare the 100% curve

and 90% curve ofFigure2-21.Notice that at rated load the 90%

curve requires more slip. The resulting drop in speed may causeproblems in the driven machine or process.

Reduced Speed -A prolonged drop in voltage may result

in the motor never reaching its nameplate rated base speed. Also,

speed regulation would be poor; since greater slip is required for

normal load changes.

Reduced Peak Torque- A 10% voltage decrease will

reduce Peak Torque (Breakdown Torque) by 19%. If the

application involves momentary load surges, there may not be

adequate Peak Torque to ride through the surge. Severe speed

fluctuations or even a complete stall may result.2-23

For example, consider a motor nameplate rated at 230

voltage (ENP) and 20A (INP). If line is increased to 245V (ELI&

the full load current (I& would be only 18.8 amps.IFL 5 27 = 2# + 4 = 1/./ mps

If line voltage is reduced to 208, then the full load current

will increase to a value of 22 amps. When determining expected

motor current, the following values are accurate within 20%. A

30 squirrel cage motor operating at 230V draws 3 amps/HP. A460V motor draws 1.5 amps/HP and a 575V motor draws 1 amp/HP

In contrast, a single (21 motor at 115V draws 10A/HP and at 230V

draws 5A/HP.

Using the effects of voltage variations on a motor can be

used as an advantage. By limiting the voltage to the motor during

starting, the current drawn and the torque produced by the motor

can be regulated. There are four different common methods for

reduced voltage starting: Autotransformer, Wye-Delta

(Star-Delta), Part Winding and Solid State.

Autotransformer reduced voltage starting is the most

commonly used method in North America. This method has three

adjustments which can be set to provide different torque and

current settings.Table2-Bshows a comparison of the different

types of starting and the corresponding reduced current and torque

of the motor.


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Outside North America, Wye-Delta or Star-Delta starting is

used most often. However, throughout the world Solid State

starting is very popular and in some areas has replaced traditional

electro-mechanical starters as the preferred method of reduced

voltage starting. The ability to make a wide range of adjustments

so that the motor and load can be matched more closely is one of

the reasons for its popularity.Table2-Bshows these adjustments.2-2

III. MOTORS A0 A@,STAB (R>,0C: RI&SIn the curves ofFigure2-24(see page2-29) the voltage (E)

remained constant while frequency varied. Since frequency (F)

appears in the denominator of the right hand term, stator currentmust vary as inverse of frequency (I/F).Figure2-24(see page

2-29) bears this out: 105% frequency produces only 95% current

and 95% frequency results in 105% current. They are inversely

proportional if applied voltage is held constant. This poses a

problem for motors operated from adjustable frequency drives.

Drives are capable of generating a broad range of output

frequencies to change a motors synchronous speed when required.

If a drives output voltage is fixed while its frequency is changed,

motor problems can quickly develop. For example, if the output

frequency is reduced from 60 Hz to 30 Hz (50%) the stator current

would double, overheating the motor. If frequency is increased

from 60 Hz to 120 Hz, current is halved and torque would suffer.

To prevent overheating at 30 Hz, the current must be reduced. To

provide adequate torque at 120 Hz the current must be increased.

RecallingEquation2-9, the only value left to manipulate is E, the

applied voltage.

If the stator applied voltage were to be reduced 50% while

the frequency is being decreased 50%, the ratio of voltage to

frequency would remain constant.

Stator current would not increase and assuming cooling is

not affected by speed, the motor would not overheat. Torque would

be unaffected, and the motor would perform properly at the

reduced speed. In reality, this is the method adjustable frequency

drives use: change the voltage with the frequency to maintain

proper current and torque.


15/27

Every AC motor has a ratio of voltage to frequency, known

as itsVolts per Hertz Ratio.As long as voltage and frequency

are held in this relationship, the motor will function properly. A

motors Volts per Hertz Ratio can be determined by its nameplate

data. For example, a motor nameplate for 460 Volts and 60 Hz has

a Volts per Hertz ratio of 460/60 or 7.6V/Hz. A motor rated for 380

Volts at 50 Hertz also has a 7.6V/Hz ratio.

These ratios indicate that for each 1 Hertz increase of

frequency, the voltage must be raised by 7.6 Volts to offset the

effects of inductive reactance. If the frequency is decreased by 1

Hertz, then the voltage must be lowered by 7.6 Volts for the same

reason.

3-l

At very low frequencies, the stator winding resistancebecomes a large portion of the overall impedance. As a result,

most of the applied voltage drops across the resistance. This

leaves very little voltage to excite the stator electromagnetic circuit

to produce torque. Recall that the stator impedance is really:

not just F, as used for illustration.

It is the R2 term that dominates below 15 Hertz of motor

operation. The logical solution to this problem is to boost the

applied voltage at very low frequencies to offset R2.

This is precisely what the adjustable frequency drive does

(see Figure 3-3).Some manufacturers call it Low Speed Boost,

Low Volts/Hertz Boost, or simply IR Compensation. Its effect is

shown in the torque curve ofFigure3-4(seepage3-4). The

torque curve is very flat, matching that of the DC shunt motor.

Some of the newest microprocessor controlled AC drives can match

this torque curve when coupled to a high efficiency squirrel cage

motor. The limiting factor becomes motor cooling at low speeds.

4$+&olta#e Boostedat o9 (re%uencies

34AC &OTA7&OTS?23+1 3+ 4(R>,0C: ?$+

Figure 3-3

3-3


16/27

When replacing line starter with an AC drive, the same

motor, wiring and conduits may be used. The drive may be

remotely mounted as far as one mile from the motor for

convenience or safety. Sometimes the customer may retain the line

starter as a back-up system. In the event of a system malfunction,

the line starter can bypass the drive to operate the motor

temporarily. Bypass capability is a requirement in some

applications such as water treatment pumps or boiler pumps.

Other drive methods (i.e. DC drives, eddy current clutches and

hydraulic drives) are much more expensive to bypass. (Always

check NEMA specifications and local codes to assure safe wiring

practices.)

Improved speed regulation is available with AC drives. A

line started squirrel cage motor will drop approximately 3% inspeed when fully loaded. By installing a tachometer on the motor,

the AC drive can actively monitor motor speed and improve speed

regulation to as high as 0.1%.

Operation A5o6eA motor rated for 60 Hz operation may be run at higher

frequencies when powered by an AC drive. The top speed depends

upon the voltage limits of the motor and its mechanical balancing.

230V and 460V motors normally employ insulation rated for 600V,

so the voltage limit is not usually a problem. An average 2 pole

industrial motor can safely exceed base speed by 25%. Many

manufacturers balance their 3 pole and 4 pole rotors to the same

speed - 25% over the 2 pole base speed. A 4 pole motor may

therefore operate up to 125% over base speed before reaching its

balance limit. A 60 Hz 4 pole motor might run up to 135 Hz,

whereas a 60 Hz 2 pole motor would reach its balance limit at 75

Hz. Both motors would run at the same RPM. Naturally, it is

sound advice to consult the motor manufacturer before exceeding

any motors base speed by more than 25%.

Constant &olta#e OperationFigure3-3 shows that at 60 Hz the AC drives output is

460 Volts, its maximum value. What happens if the output

frequency is increased above 60 Hz while the voltage remains at

460V? The drives Volts per Hertz ratio no longer remains constant

and available torque decreases.

If output frequency is increased to 120 Hz with 100%


17/27

voltage applied to the motor, the Volts per Hertz of the drive is no

longer 7.6 but rather 3.83. The same Volts per Hertz ratio results

when a line started motor is operated at 60 Hz with only 50%

voltage applied (for reduced voltage starting). As might be

expected, the effect on torque is the same. Recall that torque

varies as the square of the applied voltage(Equation2-4).As

such, maximum motor torque at 120 Hz is only 25% of the

maximum torque at 60 Hz.3-6

If AC drive output frequency is reduced from 120 Hz to 90

Hz at constant voltage, the Volts per Hertz ratio improves from

3.83 to 5.1 V/Hz. This is the same as providing 66% voltage at 60

Hz to a line-started motor. Torque will be 0.662 or 44% of the full

voltage torque at 60 Hz.Figure 3-5illustrates the peak torque

curve for constant voltage operation from base speed to 4 timesbase. On a 60 Hz rated motor this would cover 60 Hz to 240 Hz.

The motors rated torque (at 100% current) and intermittent

torque (at 150% current) both decay at the same rate.8'ATOR>,100

75

50

25

1.0

.06

$

Base 1.25 1.5 1.75 2 2.25 2.5 2.75 3.0 3.25 3.5 3.75 4.0

For 60Hz Motor: 60 75 90 105 120 135 150 165 180 195210 225 240

M,TI'S O( BAS S'Figure 3-5

Since the voltage, in reality, is not changing above base

speed, it is more appropriate to define torque in terms of frequency

change instead of voltage change. It can be stated then that

torque above base speed drops as the square of the frequency -

doubling the frequency, quarters the available torque. Applied

frequency and synchronous speed are equivalent, so going one step

further, torque may be defined in terms of speed. In the constant

voltage range then, motor torque drops off as the inverse of

synchronous speed squared, or 1/N2. This is shown by the curve in

Figure 3-5.

Many machine applications are constant horsepower in

their load characteristics. As speed increases, the torque required

decreases as the inverse of speed, or l/N. This torque drop-off is


18/27

not as severe as the motors l/N2 torque drop-off.Figure3-6(see

page3-8) displays the motor maximum torque and a nameplate

rated constant horsepower load.3-, O6ersie Motor

MotorBase 1.25 1.5 1.75 260 75 90 105 120 Hz

Figure 3-6

Notice that motor available torque exceeds load torque up

to approximately twice base speed. Beyond this point the load

exceeds the motor capability and would cause a stall condition.

If frequency is reduced so that the motor runs between

points A and B on the curve, a small load surge could still stall the

motor. Beyond point A there is not enough overload cushion

between the load demand and the motors peak capability. The

motor is not capable of safely powering a constant horsepower loadbeyond point A on the curve.

In the constant voltage mode of operation the range of

constant horsepower extends only to 1.5xbase speed. This does

not compare favorably with DC drives that provide constant

horsepower up to 3 or 4xbase speed. However, Voltage Source

AC Drives can achieve constant horsepower without modification

to the drive or motor - its a standard feature.

In order to exceed point A on the curve ofFigure 3-6,a

safety margin must be added to handle temporary load surges. If

the existing motor is replaced by one of greater horsepower, the

motors peak curve will increase, providing a safety margin as

shown in the insert ofFigure 3-6.However, with the existing

motor, the curve cannot be raised. Instead, the load must be

reduced to provide a margin for overloads.

By lowering the load curve to 2/3 of the motors peak curve,

an acceptable overload cushion is established.Figure3-7shows

the derated curve for operation above 1.5xbase speed. Derating

assures an adequate cushion for momentary overloads.

3-)The method of motor cooling will affect available torque in

speed ranges above1.5to 1. Shaft mounted cooling fans present a

variable torque load to the motor, which becomes appreciable at

high speeds. Other factors, such as the exact load profile, the

motors actual peak torque, service factor, and environment act to

complicate matters. Often times it will be necessary to consult the


19/27

motor manufacturer and AC drive application engineer before

exceeding the 1.5 to 1 speed range.Motor'ea"1.2 1.4 1. 1. 2.+ 2.2M,TI'S O( BAS S'

Figure 3- 7

Some motor applications, such as conveyors, require 100%

(nameplate) torque throughout the speed range. To increase

production it may be desired to operate the conveyor motor above

its base speed. An AC drive in the constant voltage mode can

provide 100% torque for a limited speed range.Figure3-8(see

page3-10) illustrates that 100% torque is available up to 1.25x

base speed while maintaining a 25% overload cushion.3-9

'ea"

TOR>,Tor%ue at Rated Current1.2 1. 2S'

Figure 3-8

Notice that the motor torque at rated (nameplate) current

drops off just as the peak torque does. In order to produce rated

torque, the motor must draw greater than rated current. The

additional heat generated is normally offset by increased airflow

from the motors fan. However, totally enclosed motors may

overheat if rated torque is demanded continuously above base

speed. Consult the motor manufacturer when employing totallyenclosed motors above rated speed. Oversize motors (higher

horsepower) may be recommended to assure adequate cooling and

torque.

Constant Tor%ue Ran#eAn AC drive in the Constant Voltage mode cannot provide

continuous constant torque beyond the 1.25 to 1 speed range

(oversizing the motor may extend this range to 1.5 to 1). To

develop Constant Torque beyond this limit, the drive must operate

in a true Constant Torque mode, not the Constant Voltage mode.

The drives Volts per Hertz must again match the motors Volts per

Hertz ratio. There are 3 ways to achieve this.

1. Rewind the Motor -Most stators are wound such that

base speed is achieved at 60 Hz or 50 Hz. Higher base speeds

(such as 90 Hz or 120 Hz) can be obtained by special request to the

manufacturer. The curves of a pole motor wound for 90 Hz base


20/27

speed are shown inFigure3-9.

3-10

AC drives do not produce sinuosidal voltages, but rather

generate square wave or pulses of voltage, which would saturate

and destroy a standard transformer. Special purpose customdesigned transformers are therefore essential for proper operation.

These transformers become quite expensive, approaching or

exceeding the price of a small AC drive. It may not be economical

to install such a transformer on small drive applications.

On larger applications output transformers are sometimes

used to match low voltage drive to a high voltage motor. This is

done in the petrochemical industry where the output voltage of a

460V AC drive may be stepped up to operate a 2300 Volt pump

motor from zero to its base speed. The price of such a transformer

was motivation for the development of high voltage semiconductors

for the output stages of AC drives.

Notice inFigure3-10(see page3-11) that the step-up

transformer halves the current while doubling voltage. To supply

100% (nameplate) current to the motor, the drive must produce

200% current at its output. The AC drive must be doubled in size:

a 10 HP drive is needed to power the 5 HP motor for constant

torque up to 120 Hz. If the range is reduced to 90 Hz, a 7l/2 HP

drive would provide rated current at the motor. This is because

the transformers winding ratio would be changed from 2 to 1 to

1.5 to 1.

Figure 3-11displays the voltage and torque curve of a 5

HP 230 volt motor when operated from an AC drive with an output

transformer.

T9ice Rated '2++TOR>,A0&OTA7;? 1 ++

'ea" Tor%ue(re%uenc/ ? $+ D 12+ Speed 2.+

Figure 3-11

The motor is rated for 60 Hz and has a V/Hz ratio of 230/60

or 3.83. The drives V/Hz ratio has been adjusted to 1.92 V/Hz

because its output voltage will be doubled by a transformer,

providing the motor with 3.83 V/Hz for constant torque operation.


21/27

3-12

At 60 Hz the motor receives 230 Volts and produces 5 HP

when fully loaded. At 120 Hz the motor voltage is 460 Volts. Since

both voltages and frequency are doubled, the V/Hz ratio remains at

3.83 and rated torque may be obtained. Increased motor losses at

the high speed and frequency will cause the motor to draw greater

than rated current to produce rated torque. However, the motor is

not over or under voltaged and will perform properly, producing 10

HP at 120 Hz. This justifies the 10 HP drive required to assure

adequate current. Still, the oversize drive and transformer

increase the price of a system significantly, normally making it

uneconomical. In addition, the step-up transformer limits the

available torque at low frequencies, limiting the range of constant

torque. This occurs because the transformer may saturate and

overheat at low frequencies, demanding that the current (andtorque) be reduced. The third method of producing constant torque

above rated speed is more practical.

3. Use a 460V Drive for a 230 Volt Motor -The curves

ofFigure 3-11may be obtained without placing a transformer

between the drive and motor if the drive can produce 460 Volts

directly at its output. The drives V/Hz ratio must be set at 3.83 to

accommodate a 230 volt motor. Also, since a 230 volt motor draws

twice the current of a 460 volt motor of equal horsepower, the AC

drive must double in size.

As current flows through the stator and rotor a counter

voltage is generated, just as in a transformer. The counter voltage

opposes the applied voltage and acts to limit motor current.

Normally the applied voltage exceeds the counter voltage and

power is supplied to the motor. However, if the counter voltage

exceeds the applied voltage, the motor will effectively become a

generator and feed power back onto the supply lines. The

generated power will be at the same voltage and frequency as the

supply and the motors power factor will be leading instead of

lagging. The magnitude of generated power is dependent on thedifference between applied voltage and the counter voltage.

During normal operation, motor synchronous speed exceeds

the running speed by the amount of slip as shown by point A in

Figure3-12(see page3-14). If the load overhauls the motor (pulls

the rotor faster than synchronous speed), slip reverses. The rotor

bars begin to cut the stator field from the opposite direction, which


22/27

reverses the polarity of rotor voltage. Rotor counter voltage then

exceeds the rotor induced voltage and produces a larger voltage of

mutual inductance in the stator windings. This voltage is the

same polarity as stator counter voltage and results in a net voltage

polarity reversal, feeding power back to the supply lines.3-13

TOR>,E3++;, , , :, , , , S'.25 .5 1.75

-1++; --2++; -

Figure 3-12

Point B inFigure 3-12illustrates that an overhauling load

has occurred; the rotor is moving faster than the stator field speed

dictates. Notice that 100% current flows but that torque is

negative and produces a retarding force that opposes theoverhauling load. It is an effective means of braking the load

while retrieving power that has been put into the mechanical load

by the motor originally. Braking of this type is called Regenerative

Braking and occurs frequently in many applications, such as

motor-generator sets and wind-powered AC generators.

Regeneration occurs in AC drives during the deceleration

process as well as during overhauling loads. If the motor is

running at 60 Hz and the frequency is quickly decreased to 50 Hz,

the rotor will suddenly be turning faster than synchronous speed

and begin regenerating. Negative torque rapidly decreases motor

speed to match the 50 Hz applied frequency. If frequency is

reduced further, regeneration will again occur. By ramping down

the applied frequency at a steady rate, an AC drive can effectively

and economically brake its load down to almost zero speed. No

mechanical brakes are required and much of the loads kinetic

energy can be retrieved.

This is especially desirable on high inertia loads, where

much energy is expanded during acceleration.

Quadrant 1 ofFigure3-13shows the motoring torqueduring forward operation when a squirrel cage motor is powered

from a variable frequency supply (AC Drive). Quadrant IV is the

generator torque for the same frequencies. Note that the

magnitude of peak generator torque equals peak motoring torque

and that regenerative torque is possible throughout the entire

speed range.


23/27

3-1

The energy efficient motor differs from a standard motor in

both design and manufacturing techniques. It is not, as some

suppose, a return to the old U frame motors of the past.

First of all, the energy efficient motor employs larger stator

conductors with higher conductivity. The rotor bars are larger as

well, to reduce the overall copper (or aluminum) loss which

comprises 55-60% of total motor losses. The motor flux density

and air gap are reduced to minimize the magnetizing current

required. Stator and rotor laminations are thinner to increase

resistance to the flow of eddy currents. More laminations are

added to the core stacks as well, producing a longer stator and

rotor for increased torque. Hysteresis losses, which are normally

20-25% of the total motor losses, are reduced by utilizing siliconsteel, instead of low carbon steel for the laminations. As a result,

the motor uses less current, has better power factor, and runs

cooler.

Since the energy efficient motor runs cooler, ventilation

requirements are reduced, allowing a smaller fan to be installed.

Windage loss (typically 5-9% of total losses) decreases and the

smaller fan runs quieter. These motors are less susceptible to

damage from impaired ventilation and operate well at higher

altitudes also. Cooler operation also increases motor life.

Insulation life is up to 4 times longer, an important fact, since

insulation break-down is the number 1 cause of motor failures.

Bearing lubrication lasts longer too, doubling the interval between

required lubrication. Cooler operation means that less burden is

placed on air conditioning equipment, (The Textile Industry uses

many squirrel cage motors in air conditioned factories.) Energy

efficient motors can also operate in higher ambient temperatures

without requiring extra cooling. A standard motor may need

additional cooling to survive in the same environment. This could

add substantially to the purchase price of the motor.Energy efficient motors are also more rugged than standard

motors, tolerating greater fluctuations in applied voltage, voltage

unbalance, and overload.

Some manufacturers line of energy efficient motors all

possess 1.15 service factors. Some can even tolerate 30-40%

overloads for prolonged periods. They are also capable of starting


24/27

higher inertia loads than standard motors because of their

increased thermal capacity. For example, manufacturers standard

50 HP motor can accelerate a maximum inertia of 597 lb ft while

an ener

57

efficient model can accelerate an inertia load as high as

764 lb ft .

Energy efficient motors tolerate non-sinusoidal waveforms

better than standard motors. This is important when the motor is

powered by an AC drive.Figure4-1shows the speed-torque

curves for an energy efficient and standard motor powered by a 6

step AC drive.-2

8OA

--w---- $!$"ResistantFCurrent

1G Ho + 1$+(R>,0C: D?

Figure 4-l

Notice that the standard motor has been derated to 85%.

This is due to the heating caused by the waveform shown in the

right corner. To provide the same performance as the energy

efficient motor, a larger standard motor is required. No derating is

required on the energy efficient motor, since it can tolerate the

disturbances caused by the 6 step waveform. Other types of AC

drives produce smoother waveforms that would not require

derating of the standard motor. One such type is the PWM (Pulse

Width Modulated) AC drive.

Energy efficient motors prove most cost effective in

applications where:

The cost of electricity is high. The higher the power rate, the

greater potential for savings.

The customer is concerned about power factor penalties and

peak demand charges.

Loads are constant, allowing accurate projection of potential

savings.

Excessive heating is expected, either from high ambient

conditions or severe duty cycle.


25/27

Running time exceeds idle time, again increasing savings.

Larger horsepower motors are involved, since they consume

more power and represent greater savings potentials.4-3

&. S'CIA MOTORS

The Wound Rotor Motor is an induction motor that permits

variable speed operation without the use of an AC drive. The

motors stator is identical to that of the standard polyphase

squirrel cage motor, but its rotor differs considerably. The rotor is

not of cast aluminum or copper bars, but rather consists of

insulated coils of wire connected in regular succession to form

definite poles (the same number as the stator poles). The ends of

these rotor windings are brouht out to slip rings mounted on the

motor shaft. Carbon brushes ride the slip rings to connect the

rotor windings to an external resistor network.Figure 5-lshows the controller, which allows adjustment

of the rotor resistance. By varying rotor resistance, the torque and

current characteristics can be changed. For example, high

resistance would produce high starting torque at low current,

similar to a NEMA D motor. As the motor accelerates, resistance

can be reduced to simulate a NEMA A motor. The result is

high starting torque, smooth acceleration, and optimum efficiency

at running speeds.JternalT1 T2 T3

on RotorFigure 5-l

Figure5-2(see page5-2) is the speed/torque curve of a

Wound Rotor Motor. It portrays the speed regulation obtained by

inserting various amounts of resistance in the rotor circuit. Notice

that with 100% resistance the rotor will slip 50% in speed when a

50% load is applied (Point A). Reducing resistance to 30%

improves speed regulation considerably (Point B).5-l

260

250

240230

220

210

200

190

180

170

160

8 150(, 140


26/27

130

OA 120TOR>, 110100

90

80

70

60

50

40

30

20

10

010 20 30 40 50 60 70 80 90 100

S' ; O( S:0CRO0O,SFigure 5-2

A means of variable speed operation is achieved with the

Wound Rotor Motor. Unfortunately, the motors efficiency

decreases in direct proportion to the speed reduction desired.

Because of this, the wound rotor motor is most often used on

cranes, hoists and elevators, where exact speed regulation andefficiency are not important. Efficiency may be improved by the

installation of a slip recovery system in the rotor circuit. Basically,

the external resistances are replaced by a solid state converter

which feeds the excess rotor power back to the supply lines.

Another type of special motor is the Reluctance

Synchronous Motor. It again uses the same stator assembly as a

standard squirrel cage motor and differs only in rotor construction.

Figure5-3shows a rotor punching used in a typical Reluctance

Synchronous Motor.-2

(luJBarriersAluminum

Figure 5-3

The rotor bars permit the motor to start as an induction

motor, with similar torque and current characteristics. Notice the

flux barriers inFigure 5-3,which guide the rotor flux to form

definite magnetic poles. At approximately 90% sync speed, the

rotor and stator fields begin to closely align, causing the rotor to

quickly accelerate to synchronous speed. Motor current increases

sharply during the transition to synchronous speed. However,

once in synchronism with the stator, the motor draws rated

current to produce rated torque without slip as shown in

Figure5-4. Speed regulation becomes dependent on the stability

of the applied frequency, which is normally excellent.TOR>,;S:0CRO0O,SS'


27/27

Figure 5-4

Reluctance Synchronous Motors are specified when speed

regulation is a primary factor, such as in the machine tool and

textile industries. Normal sizes range from fractional to 15 HP,

although higher horsepowers are available. Efficiencies are low

(3575%) and power factor is also poor (.45-.63). They are not

recommended for high inertia loads or heavy cyclical loads. This is

because the rotor must accelerate into its synchronous position

rapidly, making the jump from 95% speed to 100% speed very

quickly. High load inertia or a pulsating load makes this difficult.-3

Publication 150-2.7 December 1998 Copyright 1998 Rockell !nternational Printe" in #$%

motor fundamentals

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